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Let $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$ be two vectors, and $\overrightarrow{\mathrm{r}}$ be a vector along the vector $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$ such that $|\overrightarrow{\mathrm{r}}|=\sqrt{74}$. If the direction of $\vec{r}$ is opposite to that of $3 \vec{a}-2 \vec{b}$, then $\overrightarrow{\mathrm{r}}=$
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The correct answer is:
$-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$
Now, $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}=3(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})-2(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
$=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Since $\overrightarrow{\mathrm{r}}$ along the vector $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$
So, $\vec{r}=x(3 \vec{a}-2 \vec{b})=x(4 \hat{i}-3 \hat{j}+7 \hat{k})$
Now, $|\vec{r}|=\sqrt{74} \Rightarrow \sqrt{16 x^2+9 x^2+49 x^2}=\sqrt{74}$
$\Rightarrow \sqrt{74} x^2=\sqrt{74} \Rightarrow x= \pm 1$
since, $\vec{r}$ is the opposite direction of $3 \vec{a}-2 \vec{b}$
So $\mathrm{x}=-1$
$\Rightarrow \overrightarrow{\mathrm{r}}=-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
Now, $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}=3(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})-2(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
$=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Since $\overrightarrow{\mathrm{r}}$ along the vector $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$
So, $\vec{r}=x(3 \vec{a}-2 \vec{b})=x(4 \hat{i}-3 \hat{j}+7 \hat{k})$
Now, $|\vec{r}|=\sqrt{74} \Rightarrow \sqrt{16 x^2+9 x^2+49 x^2}=\sqrt{74}$
$\Rightarrow \sqrt{74} x^2=\sqrt{74} \Rightarrow x= \pm 1$
since, $\vec{r}$ is the opposite direction of $3 \vec{a}-2 \vec{b}$
So $\mathrm{x}=-1$
$\Rightarrow \overrightarrow{\mathrm{r}}=-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
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